{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261013-advance-20261023-r03-subgroup-homomorphism-0022",
  "pilot_number": "R0020-10997",
  "number": 10997,
  "family_id": "qf-mod23-subgroup-subset-closure",
  "title": {
    "zh-Hant": "第 10997 題：有限子群與同態",
    "en": "Question 10997: Product Closure of a Nonempty Subset in a Qualified Finite Group"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261013-advance-20261023-r03-subgroup-homomorphism-0022",
    "batch_id": "batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-subgroup-homomorphism"
    ],
    "title_zh": "有限子群與同態",
    "stem_zh": "完整有限群Cayley表。操作=subgroup-subset-closure；資料={\"n\":6,\"table\":[[2,5,3,0,1,4],[5,3,4,1,2,0],[3,4,0,2,5,1],[0,1,2,3,4,5],[1,2,5,4,0,3],[4,0,1,5,3,2]],\"subset\":[0,2,3]}。請依 [n,subset,closed] 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(6, [0, 2, 3], 是)",
      "explanation_zh": "現代橋梁核對=(6, [0, 2, 3], 是)；依完整明示資料重算；完整明示的有限／精確切片不自動升格為一般無限結論；同態不自動是單射；只分類指定有限群。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 3,
      "skills": [
        "有限子群與同態"
      ]
    },
    "variant": {
      "family_id": "qf-mod23-subgroup-subset-closure",
      "seed": 202610233022,
      "parameters": {
        "n": 6,
        "table": [
          [
            2,
            5,
            3,
            0,
            1,
            4
          ],
          [
            5,
            3,
            4,
            1,
            2,
            0
          ],
          [
            3,
            4,
            0,
            2,
            5,
            1
          ],
          [
            0,
            1,
            2,
            3,
            4,
            5
          ],
          [
            1,
            2,
            5,
            4,
            0,
            3
          ],
          [
            4,
            0,
            1,
            5,
            3,
            2
          ]
        ],
        "subset": [
          0,
          2,
          3
        ],
        "domain": "Complete_finite_group_morphisms",
        "operation": "subgroup-subset-closure",
        "semantic_template": "qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,3],\"table\":[[2,5,3,0,1,4],[5,3,4,1,2,0],[3,4,0,2,5,1],[0,1,2,3,4,5],[1,2,5,4,0,3],[4,0,1,5,3,2]]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "產生端候選須由主端重算；不構成人工審定、一般無限定理、Markov／可微聲明、Lean執行或形式Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-13",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261013-advance-20261023-r03-subgroup-homomorphism-0022",
    "batch_id": "batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "e241d2d5dd52e27dbacaeda3927e5d79279e5cb89a845ba155540b6307ed39c8",
      "family_id": "qf-mod23-subgroup-subset-closure",
      "seed": 202610233022,
      "semantic_template": "qf-mod23-subgroup-subset-closure/{\"n\":6,\"subset\":[0,2,3],\"table\":[[2,5,3,0,1,4],[5,3,4,1,2,0],[3,4,0,2,5,1],[0,1,2,3,4,5],[1,2,5,4,0,3],[4,0,1,5,3,2]]}"
    },
    "fields": {
      "title": "Product Closure of a Nonempty Subset in a Qualified Finite Group",
      "stem": "Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 5, 3, 0, 1, 4], [5, 3, 4, 1, 2, 0], [3, 4, 0, 2, 5, 1], [0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [4, 0, 1, 5, 3, 2]], nonempty candidate subset=[0, 2, 3] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?).",
      "answer": {
        "kind": "expression",
        "value": "(6, [0, 2, 3], Yes)",
        "explanation": "Enumerate every pair from the subset. This requested output records closure, whereas the separate full subgroup audit explicitly records identity and inverse obligations too. This explicit finite/exact slice does not automatically become a general infinite conclusion. The exact result, retaining the requested order, is (6, [0, 2, 3], Yes)."
      },
      "skills": [
        "Product Closure of a Nonempty Subset in a Qualified Finite Group"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "完整有限群Cayley表。操作=subgroup-subset-closure；資料={\"n\":6,\"table\":[[2,5,3,0,1,4],[5,3,4,1,2,0],[3,4,0,2,5,1],[0,1,2,3,4,5],[1,2,5,4,0,3],[4,0,1,5,3,2]],\"subset\":[0,2,3]}。請依 [n,subset,closed] 欄序回答。",
        "en": "Use complete qualified finite group Cayley tables with carrier labels 0 through size−1 and the given row/column order. Every group axiom is a premise already checked on the whole table. Results concern these finite groups; a homomorphism need not be injective or an isomorphism. size=6, table=[[2, 5, 3, 0, 1, 4], [5, 3, 4, 1, 2, 0], [3, 4, 0, 2, 5, 1], [0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [4, 0, 1, 5, 3, 2]], nonempty candidate subset=[0, 2, 3] normalized to sorted distinct labels. Give (size,subset,all products stay in the subset?)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "group=ambient;carrier Fin(6);complete Cayley table=[[2, 5, 3, 0, 1, 4], [5, 3, 4, 1, 2, 0], [3, 4, 0, 2, 5, 1], [0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [4, 0, 1, 5, 3, 2]]",
      "after": "typed operation closure=是;all associativity failures=[]",
      "relation": "implies",
      "rule_id": "complete_group_table_closure_associativity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "整張表先合型並核全部結合律三元組，不能只驗候選子集或同態碰到的格；此題要求已合格有限群。",
        "en": "Type the complete operation table and check every associativity triple,not only cells used by a subset or map.These tasks require qualified finite groups."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "group=ambient;candidate identity=0;left products row=[2, 5, 3, 0, 1, 4];right products column=[2, 5, 3, 0, 1, 4]",
      "after": "both sides equal carrier labels=否",
      "relation": "implies",
      "rule_id": "derive_two_sided_group_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "左右乘都須對每個x還原x，不能只看對角或預設單位元標籤0。",
        "en": "Both left and right multiplication must return every x.Do not infer identity from the diagonal or assume label0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "group=ambient;candidate identity=1;left products row=[5, 3, 4, 1, 2, 0];right products column=[5, 3, 4, 1, 2, 0]",
      "after": "both sides equal carrier labels=否",
      "relation": "implies",
      "rule_id": "derive_two_sided_group_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "左右乘都須對每個x還原x，不能只看對角或預設單位元標籤0。",
        "en": "Both left and right multiplication must return every x.Do not infer identity from the diagonal or assume label0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "group=ambient;candidate identity=2;left products row=[3, 4, 0, 2, 5, 1];right products column=[3, 4, 0, 2, 5, 1]",
      "after": "both sides equal carrier labels=否",
      "relation": "implies",
      "rule_id": "derive_two_sided_group_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "左右乘都須對每個x還原x，不能只看對角或預設單位元標籤0。",
        "en": "Both left and right multiplication must return every x.Do not infer identity from the diagonal or assume label0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "group=ambient;candidate identity=3;left products row=[0, 1, 2, 3, 4, 5];right products column=[0, 1, 2, 3, 4, 5]",
      "after": "both sides equal carrier labels=是",
      "relation": "implies",
      "rule_id": "derive_two_sided_group_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "左右乘都須對每個x還原x，不能只看對角或預設單位元標籤0。",
        "en": "Both left and right multiplication must return every x.Do not infer identity from the diagonal or assume label0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "group=ambient;candidate identity=4;left products row=[1, 2, 5, 4, 0, 3];right products column=[1, 2, 5, 4, 0, 3]",
      "after": "both sides equal carrier labels=否",
      "relation": "implies",
      "rule_id": "derive_two_sided_group_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "左右乘都須對每個x還原x，不能只看對角或預設單位元標籤0。",
        "en": "Both left and right multiplication must return every x.Do not infer identity from the diagonal or assume label0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "group=ambient;candidate identity=5;left products row=[4, 0, 1, 5, 3, 2];right products column=[4, 0, 1, 5, 3, 2]",
      "after": "both sides equal carrier labels=否",
      "relation": "implies",
      "rule_id": "derive_two_sided_group_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "左右乘都須對每個x還原x，不能只看對角或預設單位元標籤0。",
        "en": "Both left and right multiplication must return every x.Do not infer identity from the diagonal or assume label0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "group=ambient;element=0;derived identity=3;two-sided inverse candidates=[2]",
      "after": "inverse=2",
      "relation": "implies",
      "rule_id": "derive_group_inverses_from_actual_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按實際單位元查xy=e且yx=e，完整群前提保證唯一；不是把某個固定標籤當所有元素的逆。",
        "en": "Find y with xy=e and yx=e using the actual identity.The group premises ensure uniqueness;no fixed label is every element's inverse."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "group=ambient;element=1;derived identity=3;two-sided inverse candidates=[1]",
      "after": "inverse=1",
      "relation": "implies",
      "rule_id": "derive_group_inverses_from_actual_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按實際單位元查xy=e且yx=e，完整群前提保證唯一；不是把某個固定標籤當所有元素的逆。",
        "en": "Find y with xy=e and yx=e using the actual identity.The group premises ensure uniqueness;no fixed label is every element's inverse."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "group=ambient;element=2;derived identity=3;two-sided inverse candidates=[0]",
      "after": "inverse=0",
      "relation": "implies",
      "rule_id": "derive_group_inverses_from_actual_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按實際單位元查xy=e且yx=e，完整群前提保證唯一；不是把某個固定標籤當所有元素的逆。",
        "en": "Find y with xy=e and yx=e using the actual identity.The group premises ensure uniqueness;no fixed label is every element's inverse."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "group=ambient;element=3;derived identity=3;two-sided inverse candidates=[3]",
      "after": "inverse=3",
      "relation": "implies",
      "rule_id": "derive_group_inverses_from_actual_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按實際單位元查xy=e且yx=e，完整群前提保證唯一；不是把某個固定標籤當所有元素的逆。",
        "en": "Find y with xy=e and yx=e using the actual identity.The group premises ensure uniqueness;no fixed label is every element's inverse."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "group=ambient;element=4;derived identity=3;two-sided inverse candidates=[5]",
      "after": "inverse=5",
      "relation": "implies",
      "rule_id": "derive_group_inverses_from_actual_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按實際單位元查xy=e且yx=e，完整群前提保證唯一；不是把某個固定標籤當所有元素的逆。",
        "en": "Find y with xy=e and yx=e using the actual identity.The group premises ensure uniqueness;no fixed label is every element's inverse."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "group=ambient;element=5;derived identity=3;two-sided inverse candidates=[4]",
      "after": "inverse=4",
      "relation": "implies",
      "rule_id": "derive_group_inverses_from_actual_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "按實際單位元查xy=e且yx=e，完整群前提保證唯一；不是把某個固定標籤當所有元素的逆。",
        "en": "Find y with xy=e and yx=e using the actual identity.The group premises ensure uniqueness;no fixed label is every element's inverse."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[0, 0];ambient product=2",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[0, 2];ambient product=3",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[0, 3];ambient product=0",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[2, 0];ambient product=3",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[2, 2];ambient product=0",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[2, 3];ambient product=2",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S020",
      "uses": [
        "G001",
        "S019"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[3, 0];ambient product=0",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[3, 2];ambient product=2",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S022",
      "uses": [
        "G001",
        "S021"
      ],
      "before": "subset=[0, 2, 3];ordered pair=[3, 3];ambient product=3",
      "after": "product stays in subset=是",
      "relation": "implies",
      "rule_id": "subset_every_ordered_product_closure",
      "conditions": [],
      "explanation": {
        "zh-Hant": "核全部候選有序對，並保留ambient表的行列角色；有限群且子集非空是本題前提。",
        "en": "Check every ordered subset pair using the ambient row/column roles.The group is finite and the candidate nonempty as required."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "nonempty finite subset=[0, 2, 3];product closed=是",
      "after": "requested field records closure in this qualified finite group",
      "relation": "implies",
      "rule_id": "finite_nonempty_closure_not_false_insufficiency_claim",
      "conditions": [],
      "explanation": {
        "zh-Hant": "本欄只輸出乘法閉合。在有限群且非空前提下，閉合確可推出子群：元素冪有限重複給單位元與逆的正冪。另一題仍明列各義務，不在這裡誤說此有限情況閉合不足。",
        "en": "This field reports product closure.In a finite group,a nonempty closed subset does imply a subgroup:repeating powers supply identity and inverses.The separate task lists each obligation explicitly;do not falsely call closure insufficient in this finite setting."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      6,
      [
        0,
        2,
        3
      ],
      true
    ]
  },
  "conclusion": {
    "canonical_value": "(6, [0, 2, 3], 是)",
    "explanation": {
      "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations",
      "batch_path": "public/data/questions/batches/batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations/questions.jsonl",
      "batch_sha256": "cda066452f21ceb2bf1e6a2ccd37f0801301520fbcf638ce2d38d9c712e7cd6d",
      "original_line": 22,
      "original_line_sha256": "6594f17e71642dd5ecf957939103d10dd434eec6e7c883f412c73c0432ccf407",
      "question_compact_sha256": "e241d2d5dd52e27dbacaeda3927e5d79279e5cb89a845ba155540b6307ed39c8",
      "english_row_compact_sha256": "dfdefa2513367c5ca29e8d1a76727e100a20e95dafb3547fb28a37fe312a09a7",
      "english_manifest_path": "public/data/i18n/editions/batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261013-advance-20261023-r03-subgroup-homomorphism-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-subgroup-homomorphism": {
        "sha256": "2121cf9c71fab6aa9bfa10d881cdffbab5367fdfebbee9381b35fefec60837c0",
        "english_sha256": "d8b3efcc34a6efe50514c12c02d1650c151d671a36a64e0511bf4a65cf744d59",
        "original_path": "public/data/mko/subgroup-homomorphism.json",
        "english_path": "public/data/i18n/editions/mko-subgroup-homomorphism.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "bd67681fffd9e29972c322d497cd945a668a794dd25e0c8cd26f1a6ce22e4dfb",
    "original_candidate_sha256": "db70aa2aa4cfac122d5c99b80a7bacabea6bb4b48ebb72cb3bad1b5f319c362b",
    "original_candidate_line_sha256": "db70aa2aa4cfac122d5c99b80a7bacabea6bb4b48ebb72cb3bad1b5f319c362b"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261026-006-p21",
    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "b4c47938fd38085d44967226cfa8337d61334455472e7307c60d3a488cef598f",
      "question_first_witness_sha256": "3c23baa97a5933bc07b44f2257017ef4e845bf1429bd6f5381410681c43b97e4"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 23,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "796211907d27456ad70db3c72e4d14def7bf834d0b9df1bc3812f935db485c0a",
      "arithmetic_steps": 0
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}